Problem: Simplify the following expression and state the condition under which the simplification is valid. $a = \dfrac{-6k^3 + 90k^2 - 324k}{-10k^2 + 80k + 90}$
First factor out the greatest common factors in the numerator and in the denominator. $ a = \dfrac {-6k(k^2 - 15k + 54)} {-10(k^2 - 8k - 9)} $ $ a = \dfrac{6k}{10} \cdot \dfrac{k^2 - 15k + 54}{k^2 - 8k - 9} $ Simplify: $ a = \dfrac{3k}{5} \cdot \dfrac{k^2 - 15k + 54}{k^2 - 8k - 9}$ Next factor the numerator and denominator. $ a = \dfrac{3k}{5} \cdot \dfrac{(k - 9)(k - 6)}{(k - 9)(k + 1)}$ Assuming $k \neq 9$ , we can cancel the $k - 9$ $ a = \dfrac{3k}{5} \cdot \dfrac{k - 6}{k + 1}$ Therefore: $ a = \dfrac{ 3k(k - 6)}{ 5(k + 1)}$, $k \neq 9$